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**Hint:**We use the midpoint formula to solve this problem and first find out the coordinates of C and their sum.

**Formula Used:**

Coordinates of the middle point G of two points $M\left( {x,y} \right)$ and $N\left( {a,b} \right)$ which lie of the same line is given by $G\left( {\dfrac{{x + a}}{2},\dfrac{{y + b}}{2}} \right)$.

**Complete step-by-step answer:**

A,B and C collinear points, as they lie on the same line.

Let the coordinate of point C be $\left( {x,y} \right)$

Distance formula between two points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ , $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $

Using the distance formula between two points $A\left( {3,4} \right)$ and $B\left( {7,7} \right)$

\[

{d_{AB}} = \sqrt {{{\left( {7 - 3} \right)}^2} + {{\left( {7 - 4} \right)}^2}} \\

{d_{AB}} = \sqrt {16 + 9} \\

{d_{AB}} = 5 \\

\]

The distance between the points A and B is 5 units. The distance between B and C is given by

$

{d_{BC}} = {d_{AC}} - {d_{AB}} \\

{d_{BC}} = 10 - 5 \\

{d_{BC}} = 5 \\

$

It means that B is the mid-point of A and C.

The coordinates of the mid-point G, lying between the two points $M\left( {x,y} \right)$ and $N\left( {a,b} \right)$ , is given by $G\left( {\dfrac{{x + a}}{2},\dfrac{{y + b}}{2}} \right)$.

Using it, find the coordinates of point B.

Coordinates of $A\left( {3,4} \right)$ and $C\left( {x,y} \right)$

$B\left( {\dfrac{{3 + x}}{2},\dfrac{{4 + y}}{2}} \right)$

But the coordinates of B is $\left( {7,7} \right)$ . Equate the X-Coordinate and Y-Coordinate of $B\left( {\dfrac{{3 + x}}{2},\dfrac{{4 + y}}{2}} \right)$ and $B\left( {7,7} \right)$ to calculate the coordinates of $C\left( {x,y} \right)$

$

\dfrac{{3 + x}}{2} = 7 \\

x = 11 \\

$

Also,

$

\dfrac{{4 + y}}{2} = 7 \\

y = 10 \\

$

The coordinates of point C is $\left( {11,10} \right)$ .

The sum of coordinates of point C,

$

S = 11 + 10 \\

S = 21 \\

$

**Thus, the sum of the coordinates of point C is $S = 21$**

**Note:**Collinear points are those points which lie on the same line. For A, B, and C to be collinear, the area of the triangle should be equal to 0.

Important concept and formulas to be remembered are

The distance between two points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ , $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $

The midpoint of two points, all lying on the same line is calculated as , $G\left( {\dfrac{{x + a}}{2},\dfrac{{y + b}}{2}} \right)$ where two points are $M\left( {x,y} \right)$ and $N\left( {a,b} \right)$.

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